How is this a telescoping series?

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SUMMARY

The discussion centers on the telescoping nature of the series $$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$, which converges to the limit of $$ \frac{1}{2} $$ as k approaches infinity. A participant initially misinterpreted the series by incorrectly splitting the term into $$ \frac{1}{(2k-1)} + \frac{1}{(2k+1)} $$, leading to confusion about the cancellation of terms. The correct approach emphasizes the importance of algebraic manipulation over intuition, clarifying that the terms do not simply add but rather require proper fraction decomposition to reveal the telescoping nature.

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arhzz
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Homework Statement
Finding the limit of the series
Relevant Equations
Telescoping series formula
Hello !

Consider this series;

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ Now if we input k to let's say 3 we get this

$$ \frac{1}{2} + \frac{1}{3} +\frac{1}{3} + \frac{1}{5} +\frac{1}{7} + \frac{1}{7} $$. I think its fair to assume that only 1/2 should remain and the others should cancel out,that is kind of the thing by the telescoping series.But I have + across the board and the fractions won't cancel out but rather add up.I am not sure how the - is supposed to come into play here,since as I've been taught yestarteday we have to sum up the elements.The only logical solution would be that within the parantheses one element has to be negative,but I don't see how we get to that.
 
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arhzz said:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$
Are you sure about that?
 
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Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
 
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Vanadium 50 said:
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
Okay so you are definitely right,these are not the same so the algebra doesn't add up.I supposed it would have to be 1st fraction - 2nd fraction but I assumed it would be + because I don't know intuition? I saw * in the denominator and simply assumed it would be +
 
Don't use intuition. Use algebra.
 
Vanadium 50 said:
Don't use intuition. Use algebra.
Yea I guess I rushed it. It all makes sense now,thanks for pointing out the mistake.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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